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class="post-meta-separator">|</span><i class="fas fa-history fa-fw post-meta-icon"></i><span class="post-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-05-13T15:10:07.130Z" title="更新于 2021-05-13 23:10:07">2021-05-13</time></span><span class="post-meta-categories"><span class="post-meta-separator">|</span><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/everyday/categories/%E7%AE%97%E6%B3%95/">算法</a></span></div><div class="meta-secondline"><span class="post-meta-separator">|</span><span class="post-meta-pv-cv"><i class="far fa-eye fa-fw post-meta-icon"></i><span class="post-meta-label">阅读量:</span><span id="busuanzi_value_page_pv"></span></span></div></div></div></header><main class="layout" id="content-inner"><div id="post"><article class="post-content" id="article-container"><h1 id="一、解码异或后的排列"><a href="#一、解码异或后的排列" class="headerlink" title="一、解码异或后的排列"></a>一、<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/decode-xored-permutation/">解码异或后的排列</a></h1><p><strong>题目：</strong></p>
<p>给你一个整数数组 perm ，它是前 n 个正整数的排列，且 n 是个 奇数 。</p>
<p>它被加密成另一个长度为 n - 1 的整数数组 encoded ，满足 encoded[i] = perm[i] XOR perm[i + 1] 。比方说，如果 perm = [1,3,2] ，那么 encoded = [2,1] 。</p>
<p>给你 encoded 数组，请你返回原始数组 perm 。题目保证答案存在且唯一。</p>
<blockquote>
<p>示例 1：</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">输入：encoded &#x3D; [3,1]</span><br><span class="line">输出：[1,2,3]</span><br><span class="line">解释：如果 perm &#x3D; [1,2,3] ，那么 encoded &#x3D; [1 XOR 2,2 XOR 3] &#x3D; [3,1]</span><br></pre></td></tr></table></figure>

<p><strong>示例 2：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：encoded &#x3D; [6,5,4,6]</span><br><span class="line">输出：[2,4,1,5,3]</span><br></pre></td></tr></table></figure>
<p><strong>提示：</strong></p>
</blockquote>
<ul>
<li><code>3 &lt;= n &lt; 105</code></li>
<li><code>n</code> 是奇数。</li>
<li><code>encoded.length == n - 1</code></li>
</ul>
<p><strong>思路：</strong></p>
<p>本题的关键就是先求得最后一个数字或者第一个数字。有一个地方需要注意：<strong>给你一个整数数组 perm ，它是前 n 个正整数的排列，且 n 是个 奇数</strong> ，这就意味着perm中的元素为1~n，且元素唯一，可以先求出所有元素异或的结果total；其次，encoded数组中，第二个元素是perm[1]^perm[2]的结果，可从第二个元素开始，与每间隔一个元素的元素进行异或。可得perm中第二个元素到最后一个元素异或的结果count。将total和count进行异或可得perm中第一个元素的值，此时perm中所有元素都可以求得。</p>
<p><strong>代码：</strong></p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="keyword">int</span>[] decode(<span class="keyword">int</span>[] encoded) &#123;</span><br><span class="line">        <span class="comment">// 滚吧</span></span><br><span class="line">        <span class="comment">// 只要获得首元素，后边的元素就都可以算出来了</span></span><br><span class="line">        <span class="keyword">int</span>[] nums = <span class="keyword">new</span> <span class="keyword">int</span>[encoded.length+<span class="number">1</span>];</span><br><span class="line">        <span class="keyword">int</span> total = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">1</span>; i &lt;= nums.length; i++)</span><br><span class="line">            total ^= i;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">int</span> count = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">1</span>; i &lt; encoded.length; i+=<span class="number">2</span>)</span><br><span class="line">            count ^= encoded[i];</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">int</span> first = count^total;</span><br><span class="line">        nums[<span class="number">0</span>] = first;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">1</span>; i &lt; nums.length; i++)</span><br><span class="line">            nums[i] = nums[i-<span class="number">1</span>]^encoded[i-<span class="number">1</span>];</span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<hr>
<h1 id="二、子数组异或查询"><a href="#二、子数组异或查询" class="headerlink" title="二、子数组异或查询"></a>二、<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/xor-queries-of-a-subarray/">子数组异或查询</a></h1><p>题目：</p>
<p>有一个正整数数组 arr，现给你一个对应的查询数组 queries，其中 queries[i] = [Li, Ri]。</p>
<p>对于每个查询 i，请你计算从 Li 到 Ri 的 XOR 值（即 arr[Li] xor arr[Li+1] xor … xor arr[Ri]）作为本次查询的结果。</p>
<p>并返回一个包含给定查询 queries 所有结果的数组。</p>
<blockquote>
<p>示例 1：</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line">输入：arr &#x3D; [1,3,4,8], queries &#x3D; [[0,1],[1,2],[0,3],[3,3]]</span><br><span class="line">输出：[2,7,14,8] </span><br><span class="line">解释：</span><br><span class="line">数组中元素的二进制表示形式是：</span><br><span class="line">1 &#x3D; 0001 </span><br><span class="line">3 &#x3D; 0011 </span><br><span class="line">4 &#x3D; 0100 </span><br><span class="line">8 &#x3D; 1000 </span><br><span class="line">查询的 XOR 值为：</span><br><span class="line">[0,1] &#x3D; 1 xor 3 &#x3D; 2 </span><br><span class="line">[1,2] &#x3D; 3 xor 4 &#x3D; 7 </span><br><span class="line">[0,3] &#x3D; 1 xor 3 xor 4 xor 8 &#x3D; 14 </span><br><span class="line">[3,3] &#x3D; 8</span><br></pre></td></tr></table></figure>

<p><strong>示例 2：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：arr &#x3D; [4,8,2,10], queries &#x3D; [[2,3],[1,3],[0,0],[0,3]]</span><br><span class="line">输出：[8,0,4,4]</span><br></pre></td></tr></table></figure>
<p>提示：</p>
</blockquote>
<ul>
<li>1 &lt;= arr.length &lt;= 3 * 10^4</li>
<li>1 &lt;= arr[i] &lt;= 10^9</li>
<li>1 &lt;= queries.length &lt;= 3 * 10^4</li>
<li>queries[i].length == 2</li>
<li>0 &lt;= queries[i][0] &lt;= queries[i][1] &lt; arr.length</li>
</ul>
<p><strong>思路</strong>：</p>
<p>法一：暴力法。遍历queries，在遍历的过程中再逐个相加。</p>
<p>法二：利用额外的数组空间value记录，value[i] 代表arr数组从从第一个元素到第i个元素的异或结果，然后res[i] = value[queries[i][0]] ^ value[queries[i][1]+1]即可。</p>
<p><strong>代码</strong>：</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 法一：暴力法</span></span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="keyword">int</span>[] xorQueries(<span class="keyword">int</span>[] arr, <span class="keyword">int</span>[][] queries) &#123;</span><br><span class="line">        <span class="keyword">int</span>[] res = <span class="keyword">new</span> <span class="keyword">int</span>[queries.length];</span><br><span class="line">        <span class="comment">// 暴力法解决</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; queries.length; i++)&#123;</span><br><span class="line">            <span class="keyword">for</span>(<span class="keyword">int</span> j = queries[i][<span class="number">0</span>]; j &lt;= queries[i][<span class="number">1</span>]; j++)</span><br><span class="line">                res[i] ^= arr[j];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line"><span class="comment">// 法二</span></span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="keyword">int</span>[] xorQueries(<span class="keyword">int</span>[] arr, <span class="keyword">int</span>[][] queries) &#123;</span><br><span class="line">        <span class="keyword">int</span>[] res = <span class="keyword">new</span> <span class="keyword">int</span>[queries.length];</span><br><span class="line">        <span class="keyword">int</span>[] value = <span class="keyword">new</span> <span class="keyword">int</span>[arr.length+<span class="number">1</span>];</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">1</span>; i &lt; value.length; i++)&#123;</span><br><span class="line">            value[i] = arr[i-<span class="number">1</span>]^value[i-<span class="number">1</span>];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; queries.length; i++)&#123;</span><br><span class="line">            <span class="keyword">if</span>(queries[i][<span class="number">0</span>] == queries[i][<span class="number">1</span>])</span><br><span class="line">                res[i] = arr[queries[i][<span class="number">0</span>]];</span><br><span class="line">            <span class="keyword">else</span></span><br><span class="line">                res[i] = value[queries[i][<span class="number">0</span>]]^value[queries[i][<span class="number">1</span>]+<span class="number">1</span>];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<hr>
<h1 id="三、常见题型分为算法、数据结构、数学。"><a href="#三、常见题型分为算法、数据结构、数学。" class="headerlink" title="三、常见题型分为算法、数据结构、数学。"></a>三、常见题型分为算法、数据结构、数学。</h1><p><strong>1、其中算法分为：</strong></p>
<ul>
<li>贪心算法；</li>
<li>指针类问题：双指针（滑动窗口）、快慢指针；</li>
<li>二分算法；</li>
<li>排序算法；</li>
<li>优先搜索算法：深度优先搜索（回溯法）、广度优先搜索；</li>
<li>动态规划：基于一维/二维类型、分割问题、子序列问题、背包问题、字符串编辑问题、股票交易；</li>
<li>分治与递归；</li>
</ul>
<p><strong>2、数据结构：</strong></p>
<ul>
<li>数组、队列、栈、优先队列、双端队列、有序集合和映射、哈希集合和映射；</li>
<li>字符串；</li>
<li>指针结构：链表、树（树的递归、层次遍历、前中后序遍历、二叉查找树、字典树）、图（二分图、拓扑排序）</li>
<li>并查集；</li>
<li>复合数据结构；</li>
</ul>
<p><strong>3、数学：</strong></p>
<ul>
<li>因数和倍数；</li>
<li>质数；</li>
<li>数字处理；</li>
<li>随机与取样；</li>
<li>位运算；</li>
</ul>
</article><div class="post-copyright"><div class="post-copyright__author"><span class="post-copyright-meta">文章作者: </span><span class="post-copyright-info"><a href="mailto:undefined">蔡哞哞</a></span></div><div class="post-copyright__type"><span class="post-copyright-meta">文章链接: </span><span class="post-copyright-info"><a href="https://caixm1025.gitee.io/2021/05/13/05-13%E7%AE%97%E6%B3%95%E5%AD%A6%E4%B9%A0/">https://caixm1025.gitee.io/2021/05/13/05-13%E7%AE%97%E6%B3%95%E5%AD%A6%E4%B9%A0/</a></span></div><div class="post-copyright__notice"><span class="post-copyright-meta">版权声明: </span><span class="post-copyright-info">本博客所有文章除特别声明外，均采用 <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/" target="_blank">CC BY-NC-SA 4.0</a> 许可协议。转载请注明来自 <a href="https://caixm1025.gitee.io" target="_blank">个人博客</a>！</span></div></div><div class="tag_share"><div class="post-meta__tag-list"><a class="post-meta__tags" href="/everyday/tags/%E4%BD%8D%E8%BF%90%E7%AE%97/">位运算</a></div><div class="post_share"><div class="social-share" data-image="/everyday/img/%E5%9B%BE%E7%89%8713.jpg" data-sites="facebook,twitter,wechat,weibo,qq"></div><link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/social-share.js/dist/css/share.min.css" media="print" onload="this.media='all'"><script src="https://cdn.jsdelivr.net/npm/social-share.js/dist/js/social-share.min.js" defer></script></div></div><nav class="pagination-post" id="pagination"><div class="prev-post pull-left"><a href="/everyday/2021/05/14/05-14%E5%AD%A6%E4%B9%A0%E7%AC%94%E8%AE%B0/"><img class="prev-cover" src="/everyday/img/%E5%9B%BE%E7%89%8714.jpg" onerror="onerror=null;src='/everyday/img/404.jpg'" alt="cover of previous post"><div class="pagination-info"><div class="label">上一篇</div><div class="prev_info">05-14学习笔记</div></div></a></div><div class="next-post pull-right"><a href="/everyday/2021/04/09/04-09%E7%AE%97%E6%B3%95%E5%AD%A6%E4%B9%A0/"><img class="next-cover" src="/everyday/img/%E5%9B%BE%E7%89%879.jpg" onerror="onerror=null;src='/everyday/img/404.jpg'" alt="cover of next post"><div class="pagination-info"><div class="label">下一篇</div><div class="next_info">04-09算法学习</div></div></a></div></nav></div><div class="aside-content" id="aside-content"><div class="sticky_layout"><div class="card-widget" id="card-toc"><div class="item-headline"><i class="fas fa-stream"></i><span>目录</span></div><div class="toc-content"><ol class="toc"><li class="toc-item toc-level-1"><a class="toc-link" href="#%E4%B8%80%E3%80%81%E8%A7%A3%E7%A0%81%E5%BC%82%E6%88%96%E5%90%8E%E7%9A%84%E6%8E%92%E5%88%97"><span class="toc-number">1.</span> <span class="toc-text">一、解码异或后的排列</span></a></li><li class="toc-item toc-level-1"><a class="toc-link" href="#%E4%BA%8C%E3%80%81%E5%AD%90%E6%95%B0%E7%BB%84%E5%BC%82%E6%88%96%E6%9F%A5%E8%AF%A2"><span class="toc-number">2.</span> <span class="toc-text">二、子数组异或查询</span></a></li><li class="toc-item toc-level-1"><a class="toc-link" href="#%E4%B8%89%E3%80%81%E5%B8%B8%E8%A7%81%E9%A2%98%E5%9E%8B%E5%88%86%E4%B8%BA%E7%AE%97%E6%B3%95%E3%80%81%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E3%80%81%E6%95%B0%E5%AD%A6%E3%80%82"><span class="toc-number">3.</span> <span class="toc-text">三、常见题型分为算法、数据结构、数学。</span></a></li></ol></div></div></div></div></main><footer id="footer" style="background-image: url(/everyday/img/%E5%9B%BE%E7%89%8713.jpg)"><div id="footer-wrap"><div class="copyright">&copy;2020 - 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